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C. Johannesson II. The Gas Laws (p. 313-322) Ch. 10 & 11 - Gases
C. Johannesson
II. The Gas
Laws
(p. 313-322)
Ch. 10 & 11 - Gases
C. Johannesson
A. Boyle’s Law
P
V
PV = k
Volume
(mL)
Pressure
(torr)
P·V
(mL·torr)
10.0 760.0 7.60 x 103
20.0 379.6 7.59 x 103
30.0 253.2 7.60 x 103
40.0 191.0 7.64 x 103
C. Johannesson
A. Boyle’s Law
The pressure and volume
of a gas are inversely
related
at constant n & T
P
V
PV = kP1 V2
P2 V1=P1V1= P2V2
C. Johannesson
A. Boyle’s Law
C. Johannesson
kT
VV
T
B. Charles’ Law
Volume
(mL)
Temperature
(K)
V/T
(mL/K)
40.0 273.2 0.146
44.0 298.2 0.148
47.7 323.2 0.148
51.3 348.2 0.147
C. Johannesson
kT
VV
T
B. Charles’ Law
The volume and absolute
temperature (K) of a gas
are directly related
at constant n & P
V1 T1
V2 T2=
C. Johannesson
B. Charles’ Law
C. Johannesson
kT
PP
T
C. Gay-Lussac’s Law
Temperature
(K)
Pressure
(torr)
P/T
(torr/K)
248 691.6 2.79
273 760.0 2.78
298 828.4 2.78
373 1,041.2 2.79
C. Johannesson
kT
PP
T
C. Gay-Lussac’s Law
The pressure and
absolute temperature (K)
of a gas are directly
related
at constant n & V
P1 T1
P2 T2=
C. Johannesson
= kPVP
T
V
T
PV
T
D. Combined Gas Law
P1V1
T1
=P2V2
T2
P1V1T2 = P2V2T1
C. Johannesson
P1V1
T1=
P2V2
T2( )
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
A gas occupies 473 cm3 at
36°C. Find its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
C. Johannesson
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
A gas occupies 100. mL at 150.
kPa. Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
P1V1
T1=
P2V2
T2( )
C. Johannesson
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
E. Gas Law Problems
A gas occupies 7.84 cm3 at 71.8
kPa & 25°C. Find its volume at STP.
P T V
COMBINED GAS LAW
P1V1
T1=
P2V2
T2( )
C. Johannesson
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1V1T2 = P2V2T1
E. Gas Law Problems
A gas’ pressure is 765 torr at
23°C. At what temperature will
the pressure be 560. torr? GAY-LUSSAC’S LAW
P T
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = - 47°C
P1V1
T1=
P2V2
T2( )
